Karnaugh Maps 1 15

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Karnaugh Maps / K-maps by ANSHAY AGARWAL

Before going through the Karnaugh maps, some terms need to beclarified. These are as follows

Literals

A literal is a single logic variable or its complement.

For example X, Y, A, Z, X, etc.

Minterms

A minterm is the product of all the literals with or without

complement involved in a logic system.

For example

AB, AB, AB, AB (for a problem containing only A and B),

ABC, ABC, ABC (for a problem containing A, B and C)

When the values of different variables are given, minterms can

be easily formed as-

If X=0, Y=0 minterm would be XY

If X=1, Y=0, Z=1 minterm would be XYZ

So, use the variable with value 1 as it is and variable with value 0

as complemented to find the minterm.


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Maxterm

A maxterm is the sum of all the literals with or without

complement involved in a logic system.

For example

A+B, A+B, A+B, A+B (for problem containing only A and B)

A+B+C, A+B+C, (for problem containing A, B and C)

When the values of different variables are given, maxterms can

be easily formed as-

If X=0, Y=0 maxterm would be X+YIf X=1, Y=0, Z=1 maxterm would be X+Y+Z

So use the variable with value 1 as complemented and variable

with value 0 as it is to find the maxterm.

Canonical expressions

A Boolean expression containing entirely of minterms or

maxterms is known as canonical expression. These are of two

types

o Sum Of Product(SOP form)

It is the sum of all the minterms that result in a true value

of the output variable.

For example

XY+XY+XYXYZ+XYZ+XYZ+XYZ

AB+AB


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o For maxterm

To represent a maxterm as shorthand notation following steps are

to be followed-

3. Write 1 for a complemented term and 0 for non-complemented

term. This will give you a binary number.

4. The shorthand notation will be a capital M with the decimal

equivalent of the binary number as subscript of M.

Eg. The minterm X+Y is represented as-

01

M1

X+Y+Zwill be represented as

X+Y+Z --> 010

So M2

Minterm expansion of an expression

Any expression can be represented using minterms. To find the

minterm expansion of an expression following steps have to be

followed-

1. Write down all the terms in the expression

2. Put X where ever a literal is missing to convert the terms to

minterms

3. Use all the combinations of Xs to find minterms

4. Remove the duplicate/repeated terms and write the terms

together.


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Lets consider an example

Write AB+C using minterms onlyFollow the above steps as-

1. AB , C

2. ABX , XXC

3. ABX ABC, ABC

XXC ABC, ABC, ABC, ABC4. ABC+ABC+ABC+ABC+ABC-----------------answer

Write AB+AC using minterms

Follow the steps as-

1. AB, AC

2. ABX, AXC

3. ABX ABC, ABCAXC ABC, ABC

4. ABC+ABC+ABC -----------------------------answer

Exercise: write the expressions using minterms-

1) A+B

2) ABC+AD

3) AC+BD


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Revisiting SOP and POS forms

Till now we have studied what the SOP and POS forms are. Now we willsee how to expand a given function in the SOP or POS form.

Lets consider a three input valued function F such that F is 1 if

number of 1 inputs is odd and 0 otherwise.

SOP formFirst make the truth table for the function.

One special thing that we will do here is to write the minterms also in a

separate column

A B C F MINTERMS

0 0 0 0 ABC

0 0 1 1 ABC

0 1 0 1 ABC

0 1 1 0 ABC

1 0 0 1 ABC

1 0 1 0 ABC

1 1 0 0 ABC

1 1 1 1 ABC

The SOP form of F is given by the sum of those minterms for which

value of F is 1.

F = ABC+ ABC+ ABC+ ABC

See how simple it is!


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Example:

-----------Find SOP form of the function F = (0, 1, 3, 6, 7)

(A function can also be represented in the above form. Here the

numbers represent the subscript of the shorthand notation ofminterms for which function has a TRUE value.)

A B C F MINTERMS

0 0 0 1 ABC

0 0 1 1 ABC

0 1 0 0 ABC

0 1 1 1 ABC

1 0 0 0 ABC1 0 1 0 ABC

1 1 0 1 ABC

1 1 1 1 ABC

The SOP form isF = ABC + ABC + ABC + ABC + ABC

-----------Find SOP form of the function F = AB + BC

Draw the truth table

A B C F MINTERMS

0 0 0 0 ABC

0 0 1 0 ABC

0 1 0 0 ABC

0 1 1 1 ABC

1 0 0 0 ABC

1 0 1 0 ABC1 1 0 1 ABC

1 1 1 1 ABC


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SOP form is-- F = ABC + ABC + ABC

POS formMake the truth table

Write the Maxterms in a separate column

A B C F MAXTERMS

0 0 0 0 A+B+C

0 0 1 1 A+B+C

0 1 0 1 A+B+C

0 1 1 0 A+B+C

1 0 0 1 A+B+C

1 0 1 0 A+B+C1 1 0 0 A+B+C

1 1 1 1 A+B+C

The POS form of F is given by the product of those maxterms for

which value of F is 0.

F = (A+B+C)(A+B+C)(A+B+C)(A+B+C)

Example:

-----------Find POS form of the function F = (2, 4, 5)

(A function can also be represented in the above form. Here the

numbers represent the subscript of the shorthand notation of

maxterms for which function has a FALSE value.


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is used for maxterms and is used for minterms)

A B C F MAXTERMS

0 0 0 1 A+B+C

0 0 1 1 A+B+C

0 1 0 0 A+B+C

0 1 1 1 A+B+C

1 0 0 0 A+B+C

1 0 1 0 A+B+C

1 1 0 1 A+B+C

1 1 1 1 A+B+C

The POS form isF = (A+B+C)(A+B+C)(A+B+C)

-----------Find POS form of the function F = AB + BC

Draw the truth tableA B C F MAXTERMS

0 0 0 0 A+B+C

0 0 1 0 A+B+C0 1 0 0 A+B+C

0 1 1 1 A+B+C

1 0 0 0 A+B+C

1 0 1 0 A+B+C

1 1 0 1 A+B+C

1 1 1 1 A+B+C

The POS form is--F = (A+B+C) (A+B+C) (A+B+C) (A+B+C) (A+B+C)

Exercise:

Find SOP and POS forms of


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F= (0, 1, 3, 6, 9, 10, 14)

F= (1, 4, 5, 7, 12, 13, 15)

F= AB + BC + ABD

Minimization of Boolean expressionsThere are two ways to minimize a Boolean expression-

1) Algebraic method

2) Karnaugh maps

Algebraic methodIn this method the Boolean expression is reduced by using the

theorems of Boolean algebra. For example

-------simplify the function F = ABC + ABC + ABC

F = AB.(C+C) + ABC (X+X=1)

= AB . 1 + ABC= AB + ABC

= A.(B+BC) (absorption law)

= A.(B + C)

= AB + AC

-------simplify the function F= ABCD + ABCD + ABCD + ABCD

F = ABC.(D+D) + ABC.(D+D)

= ABC + ABC


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= AC.(B+B)

= AC

Karnaugh MapsKarnaugh maps also known as K-maps make the simplification of the

expressions very easy. A K-map consists of 2ncells, where n is the

number of variables in the expression which is to be reduced.

The reduction of SOP and POS forms using K-maps is shown as

follows

SOP reduction using K-maps

Each cell of the K-map represents a minterm. A K-map for two

variables is shown below

X XX

Y [0] [1]

Y

Y

The number of

cells is 4 since number

of variables is

2. The 2 variables are

X and Y written with the line. The 0 and 1 written in the square

[0]

0 1

[1]

[0]

2 3


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brackets are the possible values of X and Y. Each cell is a combination

of the value of that row and column and the number on the top-right

corner of the cell represents the decimal value of that combination.

For example- the first cell has a combination [0][0] whose decimal

equivalent is 0. Also cell number 2 has a combination [0][1] and decimal

value is 1, so the value on the top-right corner of cell 2 is 1.

K-map for 3 variables is as shown

YZ YZ YZ YZ

YZ

X [00] [01] [11] [10]

X [0]

0 1 3 2

X [1][0] ]

4 5 7 6

These are combinations for YZ

The combinations are such that only one variable changes its value in

adjacent columns.

The content of a cell is determined by the value of expression for the

value of input corresponding to the values of that row and column.


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For example if the value of expression for the input 001 is 1 then the

second cell(that is cell with row value [0] and column value [01]) will be

filled with 1.

Example: let the truth table of an expression is-

X Y Z F

0 0 0 1

0 0 1 1

0 1 0 0

0 1 1 11 0 0 0

1 0 1 0

1 1 0 1

1 1 1 1

So the K-map would be-

YZ

X YZYZYZ YZ

[00] [01] [11] [10]

X [0]

0 1 3 2

1 1 1 0

X [1][0] ]

4 5 7 6

0 0 1 1


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See how the K-map is filled for the values of F from truth table.

K-map for 4 variables

CDAB

CD CD CDCD

[10][11]

0]

[01][00]

B [00]

0 1 3 2

1 0 1 1

B [01]

4 5 7 6

1 0 1 1

B [11]

12 13 15 14

1 1 1 1

B [10]

8 9 11 10

1 1 1 1

Reduction from a K-map

Pair

A combination of two ones is called a pair. In above figure, a pair is

shown with yellow background. A pair is reduced by writing all the

variables except for those which change from complemented to un-

complemented or vice versa.

For example-

In the above map, the variable B changes from B to B.


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So, the reduced value of the pair is ACD.

A pair removes only one variable.

Quad

A combination of four ones is called a quad. In above figure, a quad is

shown with green background. A quad is also reduced by writing all the



For example- in the above map, the variable B changes from B to B

and D changes from D to D.

So, the reduced value of the quad is AC since A and C do not change

values for the four cells.

A quad removes two variables.

Octet

A combination of 8 ones is called an octet. In above figure, an octet is

shown with red background. An octet is also reduced by writing all the



For example- in the above map, the variables B, C, and D change but A

does not change.So, the reduced value of the octet is A.

An octet removes three variables.

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